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the motion of a particle in a straight line is defined by the relation x=t^4-12t^2-40

where x is in metres and t is in sec.determine the position x,velocity v, and acceleration

a of the particle at t=2sec

prince parihar , 12 Years ago
Grade 11
anser 3 Answers
Khandavalli Satya Srikanth

At t=2sec position x=(2^4-12*2^2-40)=-80 (-ve sign indicates particle moving in -ve x direction)

velocity v=dx/dt=4*t^3-24*t

at t=2sec v=4*2^3-24*2=-16m/sec

acceleration=dv/dt=12*t^2-24

at t=2sec a=12*2^2-24=24m/sec^2

Last Activity: 12 Years ago
girijesh tripathi

posiposition-68m    v--     -16m/s   accl--- + 24m/s2

Last Activity: 12 Years ago
pranjal sharma

velocity= -80m/s

accleration = - 8m/s2

posiion= -72

Last Activity: 12 Years ago
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